Originally posted by phgao
Can you provide proof? And what about the 2nd Q?
hmmm...actually, i think i made a mistake earlier. I think ABC should be 40 for the first problem.
My proof of the first question (I think this is right):
We are given that AC = XC. Therefore, it follows (law of sines, for example) that CAX = AXC. (Here my notation is that CAX refers to angle CAX)
So let CAX = AXC = x.
Also, let the radius of the circle be R.
It follows from the law of sines that XC/[sin(x)] = 2R = XC/[sin(XYC)].
It also follows from law of sines that AY/[sin(x+40)] = 2R = AY/[sin(ACY)].
We conclude from these two that:
sin(x) = sin(XYC) and
sin(x+40) = sin(ACY).
This does not mean that x = XYC and x+40 = ACY. In fact, these cannot both hold because then ABC would be 0 which doesn't make sense.
So the only solution I see that is in keeping with YXC = 40 is
XYC = 180-x and ACY = x+40. (for x < 90).
Then to find x, we just set the sum of angles within the circle to 360:
XAC + ACY + XYC + YXA = 360
-> x + (x+40) + (180-x) + (x+40) = 360
-> x = 50
Then it follows that ABC = 180 - XAC - ACY
-> ABC = 180 - (X) - (X+40) = 180 - 50 - 90 = 40
For the second problem, I was getting confused...I kept getting that the problem was not constrained, but I take it that it is, so I need to look at that problem again...